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Question: If the roots of \({{x}^{3}}+3{{x}^{2}}+4x-11=0\) are \(a\),\(b\) and \(c\) and the roots \({{x}^{3}}...

If the roots of x3+3x2+4x11=0{{x}^{3}}+3{{x}^{2}}+4x-11=0 are aa,bb and cc and the roots x3+rx2+sxt=0{{x}^{3}}+r{{x}^{2}}+sx-t=0 are a+ba+b,b+cb+c and c+ac+a, then the value of t is
A.18
B.23
C.15
D.-17

Explanation

Solution

Here we have to find the value of tt. We will first find the value of sum of roots , product of roots and sum of product of roots of first cubic equation. Then we will find only of product of roots of second cubic equation which is equal to t-t here, after simplifying the product of roots, we will get that product of roots of second cubic equation in terms of the sum of roots , product of roots and sum of product of roots of first cubic equation. We will then substitute those values to get the value of tt.

Complete step-by-step answer:
The first cubic equation is:-
x3+3x2+4x11=0{{x}^{3}}+3{{x}^{2}}+4x-11=0
The roots of this cubic equation areaa,bb andcc.
Now, we will find the sum of roots.
a+b+c=coefficient of x2coefficient of x3a+b+c=-\dfrac{\text{coefficient of }{{x}^{2}}}{\text{coefficient of }{{x}^{3}}}
Putting values, we get
\Rightarrow a+b+c=31=3............(1)a+b+c=-\dfrac{3}{1}=-3............\left( 1 \right)
Now, we will find the sum of the product of roots.
ab+bc+ca=coefficient of xcoefficient of x3ab+bc+ca=\dfrac{\text{coefficient of }x}{\text{coefficient of }{{x}^{3}}}
Putting values, we get
\Rightarrow ab+bc+ca=41=4............(2)ab+bc+ca=\dfrac{4}{1}=4............\left( 2 \right)
Now, we will find the product of roots.
abc=constant termcoefficient of x3abc=-\dfrac{\text{constant term}}{\text{coefficient of }{{x}^{3}}}
Putting values, we get
\Rightarrow abc=111=11...........(3)abc=-\dfrac{-11}{1}=11...........\left( 3 \right)
The first cubic equation is:-
x3+rx2+sxt=0{{x}^{3}}+r{{x}^{2}}+sx-t=0
The roots of this cubic equation are a+ba+b,b+cb+c and c+ac+a.
Since, we need the value of tt, we will only calculate the product of roots.
Now, we will find the product of roots.
(a+b)(b+c)(c+a)=constant termcoefficient of x3\left( a+b \right)\left( b+c \right)\left( c+a \right)=-\dfrac{\text{constant term}}{\text{coefficient of }{{x}^{3}}}
Putting values, we get
\Rightarrow (a+b)(b+c)(c+a)=t1=t\left( a+b \right)\left( b+c \right)\left( c+a \right)=-\dfrac{t}{1}=-t
Now, we will multiply all the terms on the left side of the equation.
\Rightarrow [a2b+a2c+abc+abc+ab2+b2c+ac2+bc2+abcabc]=t1=t\left[ {{a}^{2}}b+{{a}^{2}}c+abc+abc+a{{b}^{2}}+{{b}^{2}}c+a{{c}^{2}}+b{{c}^{2}}+abc-abc \right]=-\dfrac{t}{1}=-t
On taking common terms out, we get
\Rightarrow [a(ab+bc+ca)+b(ab+bc+ca)+c(ab+bc+ca)abc]=t\left[ a\left( ab+bc+ca \right)+b\left( ab+bc+ca \right)+c\left( ab+bc+ca \right)-abc \right]=-t
On further simplification, we get
\Rightarrow [(a+b+c)(ab+bc+ca)abc]=t\left[ \left( a+b+c \right)\left( ab+bc+ca \right)-abc \right]=-t
Now, we will substitute the value of (a+b+c)\left( a+b+c \right), (ab+bc+ca)\left( ab+bc+ca \right) and abcabc obtained in equation 1, equation 2 and equation 3 respectively.
\Rightarrow 3×411 -3\times 4-11=-t
\Rightarrow t=23 t=23
Thus, the correct option is B.

Note: We need to remember the relation of the roots with coefficients of the equation to solve such problems. The roots of an equation are actually the values of x which we get after solving the equation.
Quadratic equation has two roots, the cubic equation has four roots and so on. Number of roots of any equation is equal to the highest power of that equation.